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A first-order formula is ${\Delta }_1$ modulo ZFC if that theory proves it equivalent, with the same free variables, both to a Sigma-one formula in set theory and to a universal unbounded quantifier block with a bounded formula in set theory as matrix. The theory used for provable equivalence matters: modulo ZF is a stronger requirement.

In the Lévy hierarchy, a bounded formula in set theory has only quantifiers of the forms $\forall x\in y$ and $\exists x\in y$. A Sigma-one formula in set theory is an existential unbounded quantifier block followed by a bounded formula in set theory, and a Pi-one formula in set theory has a universal unbounded block. A formula is a Delta-one formula modulo ZFC if ZFC proves it equivalent, with the same free variables, both to a ${\Sigma }_1$ formula and to a ${\Pi }_1$ formula. Equivalence modulo the specified theory is part of the definition; it is not necessary that the original string have both syntactic forms. Such formulas satisfy the usual Delta-one absoluteness between transitive models satisfying the relevant axioms. The modulo-ZF analogue is a Delta-one formula in set theory.

A syntactically bounded formula in set theory is built from the atomic first-order formulas $u=v$ and $u\in v$ using Boolean operations and bounded quantifiers $\exists u\in v$ and $\forall u\in v$. These abbreviate $\exists u(u\in v\wedge \cdots )$ and $\forall u(u\in v\Rightarrow \cdots )$, with $u$ different from the bounding variable $v$.

The superscript in the question allows provable equivalence: a ZF-equivalent bounded formula is any first-order formula $\phi (\vec{v})$ for which some syntactically bounded formula in set theory $\psi (\vec{v})$ satisfiesThus ${\Delta }_0^{\mathrm{ZF}}$ is the class of first-order formulas equivalent over ZF to a syntactic ${\Delta }_0$ first-order formula. If a convention uses ${\Delta }_0^{\mathrm{ZF}}$ just for syntactically bounded set-theoretic first-order formulas, it is the smaller syntactic class; the subsequent set-theoretic absoluteness proof works for both conventions.

The essential fact is absoluteness of cardinalhood in limit ranks. Let $\alpha \le \beta$ be limit ordinals and $x\in V_{\alpha }$. Being an ordinal is absolute between these transitive sets, so a non-ordinal is a cardinal number in neither structure. For an ordinal $\delta \in V_{\alpha }$, we have $\delta <\alpha$.

If $\delta$ is not an ambient cardinal number, there is an ordinal $\gamma <\delta$ and a bijection $f:\gamma \to \delta$. With Kuratowski ordered pairs, its graph has rank of a set at most $\delta +3$. Since $\alpha$ is a limit ordinal, $\delta +3<\alpha$, so $f\in V_{\alpha }$. The assertion that $f$ is such a bijection is a bounded formula in set theory and is therefore evaluated correctly by each rank. Conversely, any internal bijection witnessing failure of cardinal number status is an actual one by the same set-theoretic absoluteness. ConsequentlyUnlike arbitrary transitive models, these rank-initial structures contain every sufficiently low-rank witness. That witness availability is what upgrades the usual downward absoluteness of cardinalhood to agreement in both directions.

Take the formula saying that a set is transitive:This is a bounded formula in set theory. If $x\in M$, transitivity of $M$ ensures that every $u\in x$ and every $v\in u$ is already in $M$, and the same holds in $N$. Both structures therefore run through exactly the same bounded witnesses and use the same actual membership relation. They agree on ${\phi }_1(x)$ for every $x\in M$, proving set-theoretic absoluteness.

Let $C(x)$ be the formula expressing cardinal number status:Ordinalhood and the assertion that a given set is a bijection between fixed sets are each expressible by a bounded formula in set theory. If $M$ regarded $x$ as a non-cardinal ordinal, a witnessing smaller ordinal and bijection would remain present and valid in $N$. Ordinalhood itself is absolute between the transitive models. Thus $N\models C(x)$ implies $M\models C(x)$: this is downward absoluteness of cardinalhood.

For failure in the other direction, put $\beta =({\omega }_1{)}^M$. This ordinal belongs to $M$, and $M\models C(\beta )$ with $\beta >\omega$. Externally $\beta$ is countable, since $\beta \subseteq M$ and $M$ is countable. By the construction of $N$, it contains a countability witness for $\beta$, and hence a bijection $\omega \to \beta$. Therefore $N\models \neg C(\beta )$. This formula is downward absolute and fails to be upward absolute.

Induct on $\alpha$ using the transitivity and inclusion already established. The zero stage is immediate. Suppose $L_{\alpha }\cap \mathrm{Ord}=\alpha$. Being an ordinal is expressible by a bounded formula in set theory, for example transitivity, transitivity of all members, and membership comparability. It is therefore absolute for the transitive set $L_{\alpha }$. The subset of its elements that are ordinals is internally definable, soBy inclusion, all smaller ordinals are present as well. Conversely, if an ordinal $\gamma$ belongs to $L_{\alpha +1}$, then $\gamma \subseteq L_{\alpha }$, since every member of a definable power set is a subset of the preceding domain. All members of $\gamma$ are ordinals, so $\gamma \subseteq L_{\alpha }\cap \mathrm{Ord}=\alpha$ and $\gamma \le \alpha$. Hence the ordinal part of the successor level is exactly $\alpha +1$.

At a nonzero limit ordinal $\lambda$,Thus

Let $\theta =M\cap \mathrm{Ord}$ be the ordinal height of a model of set theory. First $\theta \ge {\omega }_1$. Indeed, if $\theta$ were countable, the internal axiom of choice would give, for every $x\in M$, a bijection in $M$ between $x$ and some ordinal below $\theta$. Such a bijection is also valid externally, making $x$ externally countable. In particular every internal rank $V_{\xi }^M$, $\xi <\theta$, would be countable. Every element of $M$ lies in one of these internal ranks, sowould be a countable union of countable sets, a contradiction. This is why an uncountable transitive set model has uncountable ordinal height.

For $\xi <\theta$, absoluteness of constructible levels gives $L_{\xi }^M=L_{\xi }$, and that level is an element of $M$. The reason is that satisfaction in a fixed set structure uses the same domain and finite formulas in both universes, so successor definitions agree; transfinite recursion then also makes limit stages agree. Transitivity consequently givesIt remains to locate a countability witness. Given an ambient countable ordinal $\alpha$, choose an injection $f:\alpha \to \omega$. Since $V=L$, it belongs to the constructible universe. Its transitive closure, together with $f$ itself, is countable. Choose a countable elementary substructure $X\prec L_{\eta }$ of a sufficiently large limit level containing this closure pointwise. The condensation lemma for the constructible universe identifies the collapse with $L_{\beta }$ for some countable $\beta$. The collapse fixes $f$, since all its hereditary members were included. Thus $f\in L_{\beta }\subseteq L_{{\omega }_1}\subseteq M$.

The ordinal $\alpha$ itself is in $M$, since $\alpha <{\omega }_1\le \theta$. Being an injection between these fixed sets is a bounded formula in set theory, so $M$ recognizes $f$ as a countability witness. This proves countable-ordinal correctness under constructibility:The use of condensation supplies a witness below the height of $M$; merely knowing that $f$ belongs somewhere to $L$ would not be sufficient.

U+D. The assertion that a given relation is a partial order quantifies only over its underlying set and checks that it is a reflexive relation, an antisymmetric relation and a transitive relation. It is therefore a bounded formula in set theory and is absolute between transitive models.

A syntactically ${\Pi }_1$ first-order formula has an unbounded universal quantifier block followed by a bounded formula in set theory. Universal quantification makes it downward absolute between transitive models. A Pi-one formula modulo ZF may have another syntax but has such an equivalent provable in ZF.

A first-order formula is ${\Pi }_1^{\mathrm{ZF}}$ if ZF proves it equivalent, with the same free variables, to a universal unbounded quantifier block followed by a bounded formula in set theory. Bounded quantifiers do not increase its level in the Lévy hierarchy. Such properties are downward absolute between appropriate transitive models of ZF.

A first-order formula is syntactically ${\Sigma }_1$ when it is an unbounded existential quantifier block followed by a bounded formula in set theory. Equivalence provable in a specified theory gives the corresponding modulo-theory notion. Existential witnesses show upward absoluteness between transitive models containing the parameters.

A first-order formula is ZF-equivalent bounded if ZF proves it equivalent, with the same free variables, to a syntactically bounded formula in set theory. It need not itself have only bounded quantifiers. Between transitive models of ZF, use the bounded equivalent and the axioms in each model to prove set-theoretic absoluteness. The syntactic bounded-formula result requires only transitivity; the provable-equivalence extension additionally requires the theory used for the equivalence.